284 CHAPTER 8 GEOMETRY FOR AMERICANS
MN; let's call thisX. Then we will need to show that
QX ·XP=MX ·XN.
5. Use the converse of Ptolemy's Theorem (8.4.30 below) to show that the four
points are concyclic. To do so, we must show that
QM ·NP+MP·NQ = QP·MN.
All of these ideas have merit, but the second and third demand many more lines to
be drawn in, and that's too confusing for now. Likewise, Ptolemy's Theorem seems
pretty complicated.
On the other hand, option #1 looks very promising, as it does seem natural that one
of the points already labeled may in fact be the center. If M, N, P, Q are concyclic, then
MN and PQ will be chords in some circle, and by the relationship between chords and
radii (8.2. 13 ), the perpendicular bisector of these chords will pass through the center
of the circle.
Notice that AB is the perpendicular bisector of MN, since MN is perpendicular to
AB (because M and N are points on the altitude perpendicular to AB), and MN is a
chord of circle i»2 (whose diameter is AB). Likewise, AC is the perpendicular bisector
ofQP.
Hence, the only viable candidate for center is A. We know that AQ = AP and
AM = AN. We have reduced the problem to showing that AM = AP. Notice that AM
and AP are each legs of right triangles inscribed in COl and i»2, respectively. Here is a
diagram showing this, eliminating some of the prior lines to avoid clutter.
A
B
The problem reduces to something quite doable. Using similar triangles, or re
membering 8.3. 10 , we have
AM^2 =AC'·AB
and